Question

Shruti travels a distance of 224 km in four parts for a total travel time of 3 hours. Her speeds in these four parts follow an arithmetic progression, and the corresponding time taken to cover these four parts follow another arithmetic progression. If she travels at a speed of 960 meters per minute for 30 minutes to cover the first part, then the distance, in meters, she travels in the fourth part is:

• A. \(72000\)
• B. \(80000\)
• C. \(86400\)
• D. \(90000\)

Hint:

When both speeds and times form arithmetic progressions, express each term in terms of the first term and common difference, then use: \[ \text{Total distance} = \sum (\text{speed} \times \text{time}) \] to solve for the unknown common difference.

Answer 1

The Correct Option is C

Approach (use the middle-term shortcut for distance): For a 4-term time AP and 4-term speed AP, write each leg's distance and use total distance $=224$ with a symmetric average, avoiding fraction juggling.

Step 1: First leg gives $v_1 = 57.6$ km/h, $t_1 = 0.5$ h. Total time $3$ h with times in AP starting at $0.5$: sum $= 4(0.5) + 6d = 3 \Rightarrow d = \tfrac16$, so times are $\tfrac36,\tfrac46,\tfrac56,\tfrac66$ h.

Step 2: Let the four speeds be $a-3p,\,a-p,\,a+p,\,a+3p$ (a symmetric AP, common difference $2p$). Matching $v_1 = a - 3p = 57.6$ keeps things tidy.

Step 3: Total distance: \[ \tfrac{3}{6}(a-3p)+\tfrac{4}{6}(a-p)+\tfrac{5}{6}(a+p)+\tfrac{6}{6}(a+3p)=224. \] Multiply by 6: $18a + (-9-4+5+18)p = 1344 \Rightarrow 18a + 10p = 1344$.

Step 4: With $a - 3p = 57.6 \Rightarrow a = 57.6 + 3p$, substitute: $18(57.6+3p) + 10p = 1344 \Rightarrow 1036.8 + 64p = 1344 \Rightarrow p = 4.8$. Then $v_4 = a + 3p = 57.6 + 6(4.8) = 86.4$ km/h.

Step 5: Distance in part 4 $= 86.4 \times 1 = 86.4$ km $= 86400$ m.

Answer: $86400$ metres.